POJ - 2420 A Star not a Tree?【爬山算法 || 模拟退火】

Time limit 1000 ms
Memory limit 65536 kB

Luke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to connect any number of computers together in a linear arrangement. Luke is particulary proud that he solved a nasty NP-complete problem in order to minimize the total cable length.
Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn’t figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.

Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won’t move his computers. He wants to minimize the total length of cable he must buy.

Input

The first line of input contains a positive integer N <= 100, the number of computers. N lines follow; each gives the (x,y) coordinates (in mm.) of a computer within the room. All coordinates are integers between 0 and 10,000.

Output

Output consists of one number, the total length of the cable segments, rounded to the nearest mm.

题目大意

在坐标系内给定n个点的坐标
求一个点使其到这n个点距离和最小

题目分析

爬山算法 x 模拟退火

假设当前记录解坐标 $(x,y)$ ，温度为 $T$
以 $(x,y)$ 为圆心， $T$ 为半径作圆
在这个圆上随机任意一点作为下一个点判断
由于温度不断下降，随即范围也会逐渐缩小

#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
typedef double dd;
#define T 10000
#define delta 0.98
#define eps 1e-8

int read()
{
    int f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=1010;
int n;
struct node{dd x,y;}rem[maxn];

dd qsum(dd x,dd y)
{
    dd res=0;
    for(int i=1;i<=n;++i)
    {
        dd tx=x-rem[i].x,ty=y-rem[i].y;
        tx*=tx; ty*=ty;
        res+=sqrt(tx+ty);
    }
    return res;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;++i)
        scanf("%lf%lf",&rem[i].x,&rem[i].y);

        dd t=T,r=delta,x=rem[1].x,y=rem[1].y;
        while(t>eps)
        {
            dd rnd=rand()%360+1;
            dd nx=x+cos(rnd)*t,ny=y+sin(rnd)*t;
            dd de=qsum(x,y)-qsum(nx,ny);
            if(de>0) x=nx,y=ny;
            t*=r;
        }
        printf("%.0lf\n",qsum(x,y));
    }
    return 0;
}